(A) Given that $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.The determinant of matrix $A$ is $|A| = \alpha^2 - (2 \times 2) = \alpha^2

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(A) Given that $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.The determinant of matrix $A$ is $|A| = \alpha^2 - (2 \times 2) = \alpha^2 - 4$.We are given that $|A^3| = 125$.Using the property of determinants $|A^n| = |A|^n$,we have $|A|^3 = 125$.Since $125 = 5^3$,we get $|A|^3 = 5^3$,which implies $|A| = 5$.Substituting the value of $|A|$,we get $\alpha^2 - 4 = 5$.$\alpha^2 = 9$.Therefore,$\alpha = \pm 3$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 125$,then $\alpha = $

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$\pm 3$
B
$\pm 2$
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$\pm 5$
D
$0$

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If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 125$,then $\alpha = $

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